# How to prove this function is bijective

• May 7th 2010, 01:11 PM
TitaniumX
How to prove this function is bijective
I'm asked to find an explicit formula for a bijection f: (-1,1) -> R

And I know this formula f(x) = x/(x^2-1) works and I've verified it using a graphing program.

But how do I prove that this function is 1-1 and onto using the definition? It seems like the x and the x^2 throws me off. Thanks
• May 7th 2010, 02:07 PM
Raoh
Quote:

Originally Posted by TitaniumX
I'm asked to find an explicit formula for a bijection f: (-1,1) -> R

And I know this formula f(x) = x/(x^2-1) works and I've verified it using a graphing program.

But how do I prove that this function is 1-1 and onto using the definition? It seems like the x and the x^2 throws me off. Thanks

i think your function is defined from $\displaystyle \mathbb{R}\setminus \left \{ -1,1 \right \}\mapsto \mathbb{R}$ (Thinking)
• May 7th 2010, 02:11 PM
TitaniumX
Quote:

Originally Posted by Raoh
i think your function is defined from $\displaystyle \mathbb{R}\setminus \left \{ -1,1 \right \}\mapsto \mathbb{R}$ (Thinking)

True, but however the function is clearly a bijection from (-1,1)->R. How does one prove something like that?
• May 7th 2010, 03:20 PM
Plato
Quote:

Originally Posted by TitaniumX
True, but however the function is clearly a bijection from (-1,1)->R. How does one prove something like that?

Suppose that $\displaystyle f:( - 1,1) \to \mathbb{R};~~x \mapsto \frac{x}{{x^2 - 1}}$.
To show that the function is injective suppose that $\displaystyle f(a)=f(b)$ then show that $\displaystyle a=b$.
Note that if $\displaystyle \frac{a}{{a^2 - 1}} = \frac{b}{{b^2 - 1}}$ then it must be the case that $\displaystyle a~\&~b$ have the same sign.
And from the equality we get $\displaystyle ab^2 - a = a^2 b - b \Rightarrow ab^2 - a^2 b = a - b \Rightarrow ab\left( {b -a} \right) = a - b$.
What is wrong with that? (Remember they have same sign)
Does that mean that $\displaystyle a=b?$

To do the surjective part, for all $\displaystyle r\in \mathbb{R}$ does $\displaystyle \frac{x}{x^2-1}=r$ have a solution in $\displaystyle (-1,1)?$
• May 7th 2010, 11:56 PM
TitaniumX
Quote:

Originally Posted by Plato
Suppose that $\displaystyle f:( - 1,1) \to \mathbb{R};~~x \mapsto \frac{x}{{x^2 - 1}}$.
To show that the function is injective suppose that $\displaystyle f(a)=f(b)$ then show that $\displaystyle a=b$.
Note that if $\displaystyle \frac{a}{{a^2 - 1}} = \frac{b}{{b^2 - 1}}$ then it must be the case that $\displaystyle a~\&~b$ have the same sign.
And from the equality we get $\displaystyle ab^2 - a = a^2 b - b \Rightarrow ab^2 - a^2 b = a - b \Rightarrow ab\left( {b -a} \right) = a - b$.
What is wrong with that? (Remember they have same sign)
Does that mean that $\displaystyle a=b?$

To do the surjective part, for all $\displaystyle r\in \mathbb{R}$ does $\displaystyle \frac{x}{x^2-1}=r$ have a solution in $\displaystyle (-1,1)?$

Also, I've proven linear functions using the definition of 1-1 and onto with ease, but it seems like this function is a lot more complicated to prove, especially the onto part where you have to show that the y is in the domain (-1,1).
• May 8th 2010, 06:02 AM
Plato
Quote:

Originally Posted by TitaniumX

Actually that is a very detailed explanation of injectivity.
So it seems to me that you need a live tutor.
Perhaps you could team up with a fellow classmate and go over the proof together.
• May 8th 2010, 07:09 AM
novice
Quote:

Originally Posted by TitaniumX

Also, I've proven linear functions using the definition of 1-1 and onto with ease, but it seems like this function is a lot more complicated to prove, especially the onto part where you have to show that the y is in the domain (-1,1).

The domain of your function consists of a point in a plane. While the function being 1-1, the point in the domain has exactly one image in the real number line. In other words, the range of the function is not equal to the codomain, which should tell you something.

Here, I want to define my own function say $\displaystyle f(x,y)=xy$. The domain consists of only one point. I would check to see whether the image is unique. See if it's$\displaystyle f(x_1,y_2)=f(x_2,y_2)$. This implies that $\displaystyle x_1y_1=x_2y_2$, which means that $\displaystyle x_1=x_2$ and $\displaystyle y_1=y_2$. This also implies that the function is 1-1. This should not be hard since it involves only plugging something into a machine, just like putting money into a vending machine and get what you wanted. If you put money into a vending machine wishing to get an item and end up getting two, then you know the machine is out of order.

For unto function, the range of the function must equal to the codomain. Now, suppose that I am the owner of the vending machine. I have all my items displayed, say 50 cans of soda, and expect a dollar for each can, but I have only customer who has only a dollar. All I could do is sell a can of soda. I have no takers for the remaining 49 cans of soda. That's bad new, which I would say, my business is not onto my expectation.

Of course, no one would pay a dollar for a can of soda. Extortion, eh?(Rofl)
• May 8th 2010, 07:30 AM
novice
Quote:

Originally Posted by novice
The domain of your function consists of a point in a plane. While the function being 1-1, the point in the domain has exactly one image in the real number line. In other words, the range of the function is not equal to the codomain, which should tell you something.

Here, I want to define my own function say $\displaystyle f(x,y)=xy$. The domain consists of only one point. I would check to see whether the image is unique. See if it's$\displaystyle f(x_1,y_2)=f(x_2,y_2)$. This implies that $\displaystyle x_1y_1=x_2y_2$, which means that $\displaystyle x_1=x_2$ and $\displaystyle y_1=y_2$. This also implies that the function is 1-1. This should not be hard since it involves only plugging something into a machine, just like putting money into a vending machine and get what you wanted. If you put money into a vending machine wishing to get an item and end up getting two, then you know the machine is out of order.

For unto function, the range of the function must equal to the codomain. Now, suppose that I am the owner of the vending machine. I have all my items displayed, say 50 cans of soda, and expect a dollar for each can, but I have only customer who has only a dollar. All I could do is sell a can of soda. I have no takers for the remaining 49 cans of soda. That's bad new, which I would say, my business is not onto my expectation.

Of course, no one would pay a dollar for a can of soda. Extortion, eh?(Rofl)

I realized the lines in blue was a poor example for a 1-1 function. In fact, it's an example of something not being a function.

The correction example of a 1-1 function is that of a machine containing distinct items for each asking price. For the first coin you drop, you get one item, and when you drop the second coin the item must not be similar to the first one you get.

Professor Plato is a very good teacher. I hope I am not making a fool out of myself. :D